Math 331 Sample Quiz # 1 Winter 2012 Do the following. SECTION 2.1 EXERCISES (page 74) Problems 1-5, 7, 9-14, 18, 20. Answers. 3. nullity(T)=0, rank(T)=2. Thus, T is 1-1 but not onto. 11. T(8,11) = (5,-3,16) 18. T(a,b)= (b,0), N(T)=R(t) = span {(0,1)}. Answers to the rest of the problems are given in the text. Some of these problems are given below. The following theorems are applied to solve most of the problems listed above. A. Theorem 2.2: Let V and W be vector spaces and T : V −→ W be linear. If β = {v1 , v2 , . . . , vn } is a basis for V, then R(T ) = span(T (β)) = span {T (v1 ), T (v2 ), . . . , T (vn )}. B. Theorem 2.3 (Dimension Theorem): Let V and W be vector spaces and T : V −→ W be linear. If V is finite-dimensional, then nullity(T) + rank(T) = dim(V). C. Theorem 2.4: Let V and W be vector spaces and T : V −→ W be linear. T is 1-1 if and only if N (T ) = {0V }. D. Theorem 2.5: Let V and W be vector spaces of equal (finite) dimension and T : V −→ W be linear. Then the following conditions are equivalent. (a) T is 1-1. (b) T is onto. (c) rank(T) = dim(V). E. Theorem 2.6: Let V and W be vector spaces over F, and suppose that {v1 , v2 , . . . , vn } is a basis for V. For w1 , w2 , . . . , wn ∈ W , there exists exactly one linear transformation T : V −→ W such that T (vi ) = wi for i = 1, 2, . . . , n. 1. Let V and W be vector spaces (over the field F). (a) Please see your notes for the reasons for the steps in the following proofs. Prove that if T : V −→ W is a linear transformation then T (0V ) = 0W . T (00V ) = 0T (0V ) = 0W . (b) Prove that T : V −→ W is linear if and only if T (cx + y) = cT (x) + T (y) for all x, y ∈ V and for all c ∈ F . (=⇒): Assume T is linear and let c ∈ F and x, y ∈ V . Then T (cx + y) = T (cx) + T (y) = cT (x) + T (y). (⇐=): Assume T (cx + y) = cT (x) + T (y) for all x, y ∈ V and for all c ∈ F . Let c ∈ F and x, y ∈ V . (1) T (x + y) = T (1x + y) = 1T (x) + T (y) = T (x) + T (y). (2) T (cx) = T (cx + 0V ) = cT (x) + T (0V ) = cT (x) + 0W = cT (x). 2. Prove that T is a linear transformation, and find bases for both N(T) and R(T). Then compute the nullity and rank of T, and verify the dimension theorem. Finally, use the appropriate theorems in this section to determine whether T is one-to-one or onto. (a) T : R3 −→ R2 defined by T (a1 , a2 , a3 ) = (a1 − a2 , a3 ). See your notes (b) The linear transformation T : R4 −→ R3 is given by T (x1 , x2 , x3 , x4 ) = (x1 +x4 , x2 +x3 +x4 , 0). First, verify that T is linear. N (T ) = {(x1 , x2 , x3 , x4 ) ∈ R4 |T ((x1 , x2 , x3 , x4 ) = (0, 0, 0)} = {(x1 , x2 , x3 , x4 ) ∈ R4 |(x1 + x4 , x2 + x3 + x4 , 0) = (0, 0, 0)} = {(x1 , x2 , x3 , x4 ) ∈ R4 |x1 + x4 = 0, x2 + x3 + x4 = 0}. Now solve the system x1 + x4 =0, x2 + x3 + x4 = 0.  1 0 0 1 The reduced row-echelon form of A =  0 1 1 1  is the matrix A itself. 0 0 0 0 Since there are two free variables, dim(N (T )) = nullity(A) = 2. So x1 +x4 = 0, x2 +x3 +x4 = 0 =⇒ x1 = −x4 and x2 = −x3 − x4 . Let x3 = 1 and x4 = 0. Then x1 = 0 and x2 = −1. Thus (0, −1, 1, 0) is a basis vector. The other basis vector is obtained by letting, x3 = 0, x4 = 1. Then x1 = −1, x2 = −1 and the second basis vector is: (−1, −1, 0, 1). A basis for N (T ) (the null space of A) is {(0, −1, 1, 0), (−1, −1, 0, 1)}. By Theorem A, since a basis for R4 is {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}. R4 = span{T ((1, 0, 0, 0), T ((0, 1, 0, 0), T (0, 0, 1, 0), T (0, 0, 0, 1)} = {(1, 0, 0), (0, 1, 0), (0, 1, 0), (1, 1, 0)}. Thus, a basis for R(T ) is {T ((1, 0, 0, 0), T ((0, 1, 0, 0), T (0, 0, 0, 1)}. Hence rank(T) = 2. Neither one-one nor onto. Now verify Theorem B. (c) T : P2 (R) −→ P3 (R) defined by T (f (x)) = xf (x) + f 0 (x). Verification that T is linear: Let c ∈ R and f (x), g(x) ∈ P2 (R). Then T (cf (x) + g(x)) = x(cf (x) + g(x)) + (cf (x) + g(x))0 = x(cf (x) + g(x)) + cf 0 (x) + g 0 (x) = c(xf (x) + f 0 (x)) + (xg(x) + g 0 (x)) = cT (f (x)) + T (g(x)). Since {1, x, x2 } is a basis for P2 (R), R(T ) = span{T (1), T (x), T (x2 )} = span{x, x2 +1, x3 +2x}. However, {x, x2 +1, x3 +2x} is linearly independent. So, a basis for R(T ) is {x, x2 +1, x3 +2x} and rank(T) = dim(R(T)=3. By Theorem B, dim(P2 (R)) = nullity(T)+rank(T). Thus, 3= nullity(T)+3. Hence, nullity(T)= 0. So, N(T)={0}. T is 1-1 but not onto. 3. Let T : R3 −→ R2 be a function. State why T is not linear. (a) T (a1 , a2 ) = (sin a1 , 0). Let a1 = T (b1 , b2 ) 6= T (a1 + b1 , a2 + b2 ). π 2 and a2 = 0; b1 = π 2, b2 = 0. Now verify that T (a1 , a2 ) + (b) T (a1 , a2 ) = (a1 + 1, a2 ). T (0, 0) 6= (0, 0) 4. Prove that there exists a linear transformation T : R2 −→ R3 such that T (1, 1) = (1, 0, 2) and T (2, 3) = (1, −1, 4). What is T (8, 11)? By Theorem E, since {(1, 1), (2, 3)} is a basis for R2 , T is linear. (8,11) = 2(1,1)+3(2,3). 5. Let V and W be vector spaces, let T : V −→ W be linear, and let {w1 , w2 , . . . , wk } be a linearly independent subset of R(T ). Prove that if S = {v1 , v2 , . . . , vk } is chosen so that T (vi ) = wi for i = 1, 2, . . . , k, then S is linearly independent. Assume ∃c1 , cc 2, . . . , ck ∈ F such that c1 v1 +c2 v2 +. . .+ck vk = 0V . Then T (c1 v1 +c2 v2 +. . .+ck vk ) = T (0V ). So, c1 w1 +c2 w2+. . .+ck wk = 0W . But {w1 , w2 , . . . , wk } is linearly independent. So, c1 = 0, c2 = 0, . . . , cn = 0. 6. Let V and W be vector spaces and T : V −→ W be linear. (a) Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W. (=⇒): Assume T is 1-1 and {v1 , v2 , . . . , vk } be linearly independent. Now c1 T (v1 )+c2 T (v2 )+ . . .+ck T (vk ) = 0W gives T (c1 v1 +c2 v2 +. . .+ck vk ) = T (0V ) =⇒ c1 v1 +c2 v2 +. . .+ck vk = 0V , since T is 1-1. But {v1 , v2 , . . . , vk } be linearly independent. So, c1 = 0, c2 = 0, . . . , cn = 0. (⇐=): Assume T carries linearly independent subsets of V onto linearly independent subsets of W. Proof By contradiction. Assume T is not 1-1. Then, by Theorem D above, N (T ) 6= {0V )}. Then N (T ) has a nonempty basis, say {v1 , v2 , . . . , vk }. Now {v1 , v2 , . . . , vk } is a linearly independent subset of V (since N(T) ⊆ V) but {T (v1 ), T (v2 ), . . . , T (vn )} = {0W } (since vi ∈ N (T ), i = 1, 2, . . . , vn ) is linearly dependent (any subset of a vector space that contains the zero vector is linearly dependent). (b) Suppose that T is one-to-one and S is a subset of V. Prove that S is linearly independent if and only if T (S) is linearly independent. Let S={{v1 , v2 , . . . , vk }. Then T (S) = {T (v1 ), T (v2 ), . . . + T (vk )}. c1 v1 + c2 v2 + . . . + ck vk = 0V ⇐⇒ c1 T (v1 ) + c2 T (v2 ) + . . . + ck T (vk ) = 0W . Now, complete the p.roof (c) Suppose β = {v1 , v2 , . . . , vn } is a basis for V and T is one-to-one and onto. Prove that T (β) = {T (v1 ), T (v2 ), . . . , T (vn )} is a basis for W. T (β) = W (since T is onto) T (β) is LI by (b) 7. Let V and W be finite-dimensional vector spaces and T : V −→ W be linear. (a) Prove that if dim(V) < dim (W), then T cannot be onto. rank(T) < dim(V) < dim (W) (b) Prove that if dim(V) > dim (W), then T cannot be one-to-one. dim(V) > dim (W) ≥ rank(t) (Definition of R(T)) =⇒ dim(V) > rank(t). By Theorem B above, nullity(T) ≥ 1. So N(T) 6= {0V }. By Theorem C, T is not 1-1 8. Give an example of a linear transformation T : R2 −→ R2 such that N (T ) = R(T ). T(a,b)= (b,0), N(T)=R(t) = span (0,1).